Inductor DCR Power Loss
P = I² × DCR
Quick answer
This calculator solves P = I² × DCR for Power Loss. With Current 2 A, DC Resistance 50 mΩ, the result is 0.2 W. It solves in 3 directions, so you can enter the values you have and get the one you need.
Calculator
Formula
Description
Every inductor has DC resistance (DCR) from the winding wire, which dissipates power as heat proportional to the square of the current. This copper loss is the dominant loss mechanism in power inductors at low frequencies. At higher frequencies, AC resistance (from skin effect and proximity effect) adds additional loss beyond DCR. Selecting a low-DCR inductor is critical for efficiency in power converters, especially at high currents. A 10 mΩ DCR inductor carrying 5 A dissipates 250 mW continuously.
Variables
- P — Power dissipated as heat (W)
- I — RMS current through the inductor (A)
- DCR — DC winding resistance (Ω)
Practical Notes
DCR increases with temperature (copper has a positive temperature coefficient of about 0.393%/°C). At 100°C, DCR is about 30% higher than at 25°C. For accurate loss estimation at operating temperature, use: DCR_hot = DCR_25 × (1 + 0.00393 × (T − 25)). Total inductor loss includes core loss (from hysteresis and eddy currents) in addition to copper loss.
Symbols and units
| Symbol | Meaning | Default unit | Accepted units |
|---|---|---|---|
| P | Power Loss | mW | µW, mW, W, kW |
| I | Current | A | pA, nA, µA, mA, A, kA |
| DCR | DC Resistance | mΩ | mΩ, Ω, kΩ, MΩ |
Step by step
- 1. What you know: Current 2 A, DC Resistance 50 mΩ
- 2. The relation: P = I² × DCR
- 3. Substitute: I = 2 A, DCR = 50 mΩ
- 4. Result: Power Loss = 0.2 W
Related quantities from the same inputs
- P = I² × R = 0.2 W
Worked examples
Minimum range
Given: Current 2 mA, DC Resistance 0.1 mΩ
Result: Power Loss = 4 × 10⁻⁴ µW
Low range
Given: Current 20 mA, DC Resistance 1 mΩ
Result: Power Loss = 0.4 µW
Typical
Given: Current 2 A, DC Resistance 50 mΩ
Result: Power Loss = 0.2 W
High range
Given: Current 100 A, DC Resistance 0.5 Ω
Result: Power Loss = 5 kW
Maximum range
Given: Current 1 kA, DC Resistance 5 Ω
Result: Power Loss = 5000 kW
Solving for each variable
Solve for Power Loss
Current 2 mA, DC Resistance 0.1 mΩ → 4 × 10⁻⁴ µW
Solve for Current
Power Loss 100 µW, DC Resistance 0.1 mΩ → 1 A
Solve for DC Resistance
Power Loss 100 µW, Current 2 mA → 25 Ω
How Power Loss changes with Current
| Current (input) | Power Loss (result) |
|---|---|
| 0.2 A | 2 mW |
| 0.5 A | 12.5 mW |
| 1 A | 50 mW |
| 1.5 A | 112.5 mW |
| 2 A | 0.2 W |
| 3 A | 0.45 W |
| 4 A | 0.8 W |
| 8 A | 3.2 W |
| 14 A | 9.8 W |
| 20 A | 20 W |
The same result in other units
| Unit | Value |
|---|---|
| µW | 200000 |
| mW | 200 |
| W | 0.2 |
| kW | 0.0002 |
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<p><a href="https://vbatpower.com/formulas/inductor-dcr-loss/">Inductor DCR Power Loss</a> Calculator by vbatpower.com</p> Frequently asked questions
How do you calculate Inductor DCR Power Loss?
Use P = I² × DCR. With Current 2 A, DC Resistance 50 mΩ, Power Loss works out to 0.2 W.
What is the formula for Inductor DCR Power Loss?
The formula is P = I² × DCR. Every inductor has DC resistance (DCR) from the winding wire, which dissipates power as heat proportional to the square of the current.
How do you find Current when you know the other values?
Rearrange for Current: with Power Loss 100 µW, DC Resistance 0.1 mΩ, Current is 1 A.
How do you find DC Resistance when you know the other values?
Rearrange for DC Resistance: with Power Loss 100 µW, Current 2 mA, DC Resistance is 25 Ω.
What units does Inductor DCR Power Loss use?
Power Loss is shown in µW, mW, W, kW. Values are held in SI base units internally, so you can mix prefixes without converting by hand.
Related Concepts
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